Past exam of the mathematics course of the University of Cambridge 2010 ia Paper 1 2C a Solution Created 2026-09-24 Updated 2026-10-14
The polar form of a complex number and the roots of unity givesince . Using the complex conjugate to compute the squared complex modulus,The residue runs through every class modulo , already with . The cosine is even, so the nine angles give five distinct values, represented by . They are distinct because cosine is strictly decreasing on and these representative angles lie in that interval. HenceEvery listed value is attained by choosing and .
Past exam of the mathematics course of the University of Cambridge 2014 ia Paper 1 1B a i Solution Created 2026-09-24 Updated 2026-10-06
Squaring first avoids unnecessary expansion: , and squaring again gives the fourth power . In the polar form of a complex number, ; multiplying its complex argument by four gives the same negative-real result.
The number being squared has polar form of a complex number . Squaring doubles its argument, giving . Hence
Differentiate the polar form of a complex number using the product rule and chain rule:A second differentiation givesThe two contributions to arise separately from differentiating and from differentiating the moving exponential factor.
For an eigenvector , the equation readsA nonzero solution requires . Write the three roots of unity in the requested polar form of a complex number:Indeed satisfies all three equations. These eigenvectors are unnormalized; any nonzero scalar multiples are equally valid.
In polar coordinates, , , with . Thus the polar form of a complex number isHere is its complex modulus; for , the complex argument is determined modulo . At the angle is undefined, although every value of gives the same point when .